What is the slope of the line tangent to $f(x) = 2x^{2}-x-8$ at $x = 2$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x+\Delta x)^{2}-(x+\Delta x)-8) - (2x^{2}-x-8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x^{2}+2x \Delta x+\Delta x^{2})-(x+\Delta x)-8) - (2x^{2}-x-8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2x^{2}+4(x \Delta x)+2\Delta x^{2}-x-\Delta x-8-2x^{2}+x+8}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{4(x \Delta x)+2\Delta x^{2}-\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} 4x+2(\Delta x)-1$ $ = 4x-1$ $ = (4)(2)-1$ $ = 7$